20

Recall the following relevant definitions. We say that

  • $b$ is divisible by $a$ in $R$, or $a\mid b$ in $R$, if $b = r a$ for some $r\in R$.

  • $a$ and $b$ are associates in $R$ if $a\mid b$ and $b\mid a$, (or, equivalently, if $aR = bR$).

  • $u\in R$ is a unit if it has a multiplicative inverse (a $v\in R$ such that $uv=vu=1$).

  • $a$ and $b$ are unit multiples in $R$ if $a = ub$ for some unit $u\in R$.

Given these definitions, my question is,

If $R$ is a commutative ring with unity and $a,b\in R$ are associates in $R$, are $a$ and $b$ unit multiples in $R$?

I was told that this not always true. But I encountered some difficulties in finding a counterexample.

azimut
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ShinyaSakai
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5 Answers5

30

See the following paper,

When are Associates Unit Multiples?
D.D. Anderson, M. Axtell, S.J. Forman, and Joe Stickles
Rocky Mountain J. Math. Volume 34, Number 3 (2004), 811-828.

It is mostly concerned with finding sufficient conditions on commutative rings that ensure that $Ra=Rb$ implies $a = bu$ for a unit $u$, but they do give some examples of $R$ where this fails. In particular this simple example of Kaplansky. Let $R=C[0,3]$, the set of continuous function from the interval $[0,3]$ to the reals. Let $f(t)$ and $g(t)$ equal $1-t$ on $[0,1]$, zero on $[1,2]$ but let $f(t)=t-2$ on $[2,3]$ and $g(t)=2-t$ on $[2,3]$. Then $f$ is not a unit multiple of $g$ in $R$ but each divides the other.

Robin Chapman
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  • Thank you very much for answering my question, but I think I am not quite clear about this answer. Here, if I want to find some $t(x)$ such that $f(x)=t(x)g(x)$, I should set $t(x)=1$ in the interval $[0,2]$, and $t(x)=-1$ in the interval $[2,3]$, but $t(2)=$?? Even if I determine the value of $t(2)$, $t(x)$ is not a continuous function, thus it does not belong to $R=C[0,3]$. Many thanks. – ShinyaSakai Dec 15 '10 at 01:49
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    I read the paper and find that $t(x)$ can be set to $1$ on $[0,1]$, $3-2t$ on $[1,2]$ and $-1$ on $[2,3]$. Now I understand. Many thanks. – ShinyaSakai Dec 15 '10 at 02:16
10

This is true if $R$ is a domain. More generally this is true if $a$ or $b$ is not a zero divisor in $R$. Suppose, $a$ is not a zero divisor and there exist $r,s\in R$ such that $a=rb$ and $b=sa$, then $a=rsa$, so $a(1-rs)=0$. Since $a$ is not a zero divisor, $1-rs=0$, so $rs=1$. So, $r,s$ are units. So $a$ and $b$ are associates.

7

On earlier math forums I often cited the little-known article below on this topic (e,g. see sci.math Oct 15, 2008 google groups or Ask an Algebraist 2008, and mathforum, etc)

Beware that the equivalence $\rm\ aR = bR \iff a$ is a unit multiple of $\,\rm b\,$ in $\rm\,R\,$ generally fails if $\rm R $ has zero-divisors; then there are at least a few different interesting notions of "associate", e.g.

  • $\ a\sim b\ $ are $ $ associates $ $ if $\, a\mid b\,$ and $\,b\mid a$
  • $\ a\approx b\ $ are $ $ strong associates $ $ if $\, a = ub\,$ for some unit $\,u\ \,$ (a.k.a. unit multiples)
  • $\ a \cong b\ $ are $ $ very strong associates $ $ if $\,a\sim b\,$ and $\,a\ne 0,\ a = rb\,\Rightarrow\, r\,$ unit

See said paper below for much further discussion. See also the survey linked here for the effect that this bifurcation has on the notion of unique factorization ring and related matters.

When are Associates Unit Multiples?
D.D. Anderson, M. Axtell, S.J. Forman, and Joe Stickles
Rocky Mountain J. Math. Volume 34, Number 3 (2004), 811-828.
https://projecteuclid.org/journals/rocky-mountain-journal-of-mathematics/volume-34/issue-3/When-are-Associates-Unit-Multiples/10.1216/rmjm/1181069828.full

Bill Dubuque
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1

For a counterexample, I will use the ring $R := \mathbb{Z}[a, b, r, s] / \langle b - ra, a - sb \rangle$. In this ring, we certainly have $b = ra$ and $a = sb$. It remains to show there is no unit $u \in R^*$ such that $b = ua$.

Thus, suppose we did have $b = ua$. Then $a(u-r) = 0$. On the other hand (and here is where the argument gets somewhat technically involved), it can be shown that the kernel of the $R$-module homomorphism $a\cdot : R \to R$ is generated by $rs - 1$. Therefore, there exists $p \in R$ such that $u = r + p(rs - 1)$. Also, if $u$ is a unit, then the image $\bar u$ of $u$ in $R / \langle a, b \rangle \simeq \mathbb{Z}[a,b,r,s] / \langle b - ar, a - sb, a, b \rangle = \mathbb{Z}[a,b,r,s] / \langle a, b \rangle \simeq \mathbb{Z}[r,s]$ would also have to be a unit. But the only units of $\mathbb{Z}[r,s]$ are the constant polynomials $\pm 1$, whereas with $\bar u = r + \bar p (rs - 1)$ we would need to have $\bar u(1, 1) = 1 \ne -1 = \bar u(-1, -1)$ if we consider $\bar u$ as inducing a function $\mathbb{Z}^2 \to \mathbb{Z}$. This gives a contradiction, so the assumption that $u$ is a unit must have been false.


The basic idea here is that we are in some sense taking the "universal example" of a ring with two associate elements $a,b$ (though with the caveat that the elements $u,v$ making the elements associate are also distinguished, rather than simply assumed to exist somewhere in the ring; and also that the actual universal example would be $\mathbb{Z}[a,b,r,s] / \langle b-ra, a-sb \rangle$ whereas I used $\mathbb{Q}$ for a slightly easier time with the technical details I omitted). If we had found a unit within $\mathbb{Q}[a,b,r,s] / \langle b-ra, a-sb \rangle$ making $a$ and $b$ associates, then that expression would have worked in particular for any $\mathbb{Q}$-algebra. On the other hand, once we instead found that there is no unit in this universal example, we already had our counterexample that we needed.

(On the other hand, this shows that trying to work directly with the universal example might not always be the best or easiest approach. In this example, we had some mildly hairy calculations, ultimately based on Groebner basis theory, that were needed. However, in the counterexample based on $R = C[0,3]$ in another answer, it was more straightforward to check the details.)


Addendum on the technical detail omitted above, verifying the fact in Macaulay2:

Macaulay2, version 1.16
with packages: ConwayPolynomials, Elimination, IntegralClosure, InverseSystems, LLLBases, MinimalPrimes, PrimaryDecomposition, ReesAlgebra, TangentCone, Truncations

i1 : R = ZZ[a,b,r,s] / (b - ra, a - sb)

o1 = R

o1 : QuotientRing

i2 : mult_a = map(R^1, R^1, matrix {{ a }})

o2 = | a |

         1       1

o2 : Matrix R <--- R

i3 : kernel(mult_a)

o3 = image | rs-1 |

                         1

o3 : R-module, submodule of R

1

Let me add this historic counterexample from I. Kaplansky: Elementary divisors and modules. Trans. Amer. Math. Soc. 66, 464-491 (page 466, example (b)).

Take the (Noetherian!) subring $R \subset \mathbb Z \times \mathbb F_5[x]$ consisting of all $(z,f)$ with $f(0) = z\bmod 5$. The elements $a = (0,x)$ and $b = (0,2x)$ are associates by $b = 2a$ and $a = 3b$. The unit group of $R$ consists only of the elements $\pm 1$, so $a$ and $b$ are no unit multiples.

azimut
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